2. Curvilinear Motion.pdf

Chinese University of Hong Kong

12-4. General Curvilinear Motion Application:

The path of motion of each plane in this formation can be tracked with radar and their x, y, and z coordinates (relative to a point on earth) recorded as a function of time. Problems: How can we determine the velocity or acceleration of each plane at any instant? Should they be the same for each aircraft? Curvilinear motion occurs when the particle Fig.12-6

moves along a curved path Department of Mechanical and Automation Engineering

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9 Position The position of the particle, measured from a fixed point O, is designated by the position vector r = r(t).

Fig.12-7

9 Displacement Suppose during a small time interval Δt, the particle moves a distance Δs along the curve to a new position P’, defined by r’ = r + Δr. The displacement Δr represents the change in the particle’s position. Department of Mechanical and Automation Engineering

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9 Velocity: represents the rate of change in the position of a particle. The average velocity of the particle during the time increment Δt is

vavg = Δr/Δt The instantaneous velocity is the timederivative of position

v = dr/dt The velocity vector, v, is always tangent to the path of motion.

Fig.12-7(c)

The magnitude of v is called the speed. Since the arc length Δs approaches the magnitude of Δr as t→0, the speed can be obtained by differentiating ds the path function

v=

dt

(12-6)

DepartmentNote of Mechanical andisAutomation Engineering that this not a vector!

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9 Acceleration: rate of change in the velocity of a particle. • If a particle’s velocity changes from v to v’ over a time increment Δt, the average acceleration during that increment is: aavg = Δv/Δt = (v - v’)/Δt

• The instantaneous acceleration is the timederivative of velocity: a = dv/dt = d2r/dt2

(12-7)

• A plot of the locus of points defined by the arrowhead of the velocity vector is called a hodograph. The acceleration vector is tangent to the hodograph, but not, in general, tangent to the path function.

Department of Mechanical and Automation Engineering

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12-5. Curvilinear Motion: Rectangular Components

9 Position: The position of the particle can be defined at any instant by the position vector

(12-8)

r = xi + yj + zk The magnitude of r is always positive and defined as

r = x2 + y2 + z 2 Here, the x, y, z components may all be functions of time, i.e., x = x(t), y = y(t), and z = z(t). The direction of r is specified by the components of the unit vector ur = r/r

Fig.12-8(a)


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9 Velocity: Time derivative of the position vector

dr v= = vx i + v y j + vz k dt where v x = x& v y = y& v z = z&

(12-9) (12-10)

The velocity has a magnitude defined as the positive value of

v = v x2 + v y2 + v z2 and a direction that is specified by the components of the unit vector uv=v/v and is always tangent to the path. Fig.12-8(b)


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9 Acceleration: time derivative of the velocity vector (second derivative of the position vector)

dv = axi + a y j + azk (12-11) dt a x = v&x = &x& (12-12) a y = v& y = &y&

a= where

a z = v&z = &z& The acceleration has a magnitude defined as the positive value of

Fig.12-8(c)

a = a x2 + a y2 + a z2

¾ The acceleration has a direction specified by the components of the unit vector ua = a/a. ¾ Since a represents the time rate of change in velocity, a will not be tangent to the path (Fig.12-8(c)). Department of Mechanical and Automation Engineering

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Example 3 The motion of two particles (A and B) is described by the position vectors

rA = [3ti + 9t(2 –t)j] m rB = [3(t2 –2t+2)i + 3(t –2)j] m Solution:

Find: The point at which the particles collide and their speeds just before the collision.

1).The particles will collide when their position vectors are equal, or rA = rB

so xA=xB and yA=yB . x-components: 3t = 3(t2 – 2t + 2) Simplifying: t2 – 3t + 2 = 0 Solving: t ={3 ± [32 – 4(1)(2)]0.5}/2(1) → t = 2 or 1 s y-components: 9t(2-t) =3(t–2) Simplifying: 3t2–5t–2 = 0 Solving: t ={5±[52 – 4(3)(–2)]0.5}/2(3) → t = 2 or –1/3 s Department of Mechanical and Automation Engineering

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So, the particles collide when t = 2 s. Substituting this value into rA or rB yields

xA = xB = 6 m

and yA = yB = 0

2). Their speeds can be determined by differentiating the position vectors. That is, differentiating rA and rB to get the velocity vectors, and then obtain the speeds which are the magnitudes of the corresponding velocity vectors.

. . vA = drA/dt = .xA i + yA j = [3i +(18 –18t)j] m/s At t = 2s: vA = [3i – 18j] m/s vB = drB/dt = x• Bi + y•Bj = [(6t – 6)i + 3j] m/s At t = 2s: vB = [6i + 3j] m/s Speed is the magnitude of the velocity vector, vA = (32 + 182)0.5 = 18.2 m/s; vB = (62 + 32)0.5 = 6.71 m/s Department of Mechanical and Automation Engineering

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Exercise: 1. The path of a particle is defined by y = 0.5x2. If the component of its velocity along the x-axis at x = 2 m is vx = 1 m/s, its velocity component along the y-axis at this position is A) 0.25 m/s B) 0.5 m/s

C) 1 m/s D) 2 m/s

2. If a particle has moved from A to B along the circular path in 4s, what is the average velocity of the particle ? y A) 2.5 i m/s R=5m x B) 2.5 i +1.25j m/s C) 1.25 π i m/s A B D) 1.25 π j m/s 3. The position of a particle is given as r = (4t2 i – 8t j) m. Determine the particle’s acceleration. A) (4 i +8 j ) m/s2 C) (8 i) m/s2

B) (8 i -16 j ) m/s2 D) (8 j ) m/s2


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12-7. Curvilinear Motion: Normal and Tangential Components Application:

Motorists traveling along a clover-leaf interchange experience a normal acceleration due to the change in direction of the velocity. Problem: If the car’s speed is increasing at a known rate as it travels along a curve, how can we determine the magnitude and direction of its total acceleration? Fig.12-9

Objective: Be able to determine the normal and tangential components of velocity and acceleration of a particle traveling along a curved path.


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Normal and Tangential Components When a particle moves along a curved path, it is sometimes convenient to describe its motion using coordinates other than Cartesian. When the path of motion is known, normal (n) and tangential (t) coordinates are often used. In the n-t coordinate system, the origin is located on the particle (the origin moves with the particle). 9The t-axis is tangent to the path (curve) at the instant considered, positive in the direction of the particle’s motion. 9The n-axis is perpendicular to the t-axis with the positive direction toward the center of curvature O’. Department of Mechanical and Automation Engineering

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The positive n and t directions are defined by the unit vectors un and ut, respectively. The center of curvature, O’, always lies on the concave side of the curve. The position of the particle at any instant is defined by the distance, s, along the curve from a fixed reference point. The radius of curvature, ρ, is defined as the perpendicular distance from the curve to the center of curvature at that point.


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Velocity in the n-t Coordinate System The velocity vector is always tangent to the path of motion (t-direction). The magnitude is determined by taking the time derivative of the path function, s(t).

where

v = vut . v = s = ds/dt

(12-13) (12-14)

Here v defines the magnitude of the velocity (speed) and ut defines the direction of the velocity vector.


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Acceleration in the n-t Coordinate System Acceleration is the time rate of change of velocity:

. . a = dv/dt = d(vut)/dt = vut + vut .

Here v represents the change in the . magnitude of velocity and ut represents the rate of change in the direction of ut.

(12-15)

After mathematical manipulation, the acceleration vector can be expressed as:

. a = vut + (v2/ρ)un = atut + anun

(12-16)

Fig.12-10(e)


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There are two components to the acceleration vector:

a = at ut + an un • The tangential component is tangent to the curve and in the direction of increasing or decreasing velocity.

. at = v

or

at ds =v dv

(12-17)

• The normal or centripetal component is always directed toward the center of curvature of the curve

an = v2/ρ

(12-18)

• The magnitude of the acceleration vector is

a =

a t2 + a n2

(12-19)


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Two Special Cases of Motion To summarize these concepts, consider the following special cases of motion

1) If the particle moves along a straight line, then

ρ → ∞ => an = v2/ρ = 0 =>

. a = at = v

The tangential component represents the time rate of change in the magnitude of the velocity. 2) The particle moves along a curve at constant speed, then

. at = v = 0 => a = an = v2/ρ The normal component represents the time rate of change in the direction of the velocity.


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Three-Dimensional Motion If a particle moves along a space curve, the n and t axes are defined as before. At any point, the t-axis is tangent to the path and the n-axis points toward the center of curvature. The plane containing the n and t axes is called the osculating plane. A third axis can be defined, called the Fig.12-11 binomial axis, b. The binomial unit vector, ub, is directed perpendicular to the osculating plane, and its sense is defined by the cross product

ub = ut × un There is no motion, thus no velocity or acceleration, in the binomial direction. Department of Mechanical and Automation Engineering

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Example 4 Given: Starting from rest, a motorboat travels around a circular path of r = 50 m at a speed that increases with time,

v = (0.2 t2) m/s. Find: The magnitudes of the boat’s velocity and acceleration at the instant t = 3s. Solution:

The boat starts from rest (v=0 when t=0). 1) Calculate the velocity at t=3s using v(t). The velocity vector is v = v ut , where the magnitude is given by v = (0.2t2) m/s. At t=3s: v = 0.2t2 = 0.2(3)2 = 1.8 m/s Fig.12-12


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2). Calculate the tangential and normal components of acceleration and then the magnitude of the acceleration vector.

.

The acceleration vector is a = atut + anun = vut + (v2/ρ)un

. Tangential component: at = v = d(0.2t2)/dt = 0.4t m/s2 At t=3s: at = 0.4t = 0.4(3) = 1.2 m/s2 Normal component: an = v2/ρ = (0.2t2)2/(ρ) m/s2

At t=3s: an = [(0.2)(32)]2/(50) = 0.0648 m/s2 The magnitude of the acceleration is

a = [(at)2 + (an)2]0.5 = [(1.2)2 + (0.0648)2]0.5 = 1.20 m/s2


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Exercise: 1. The normal component of acceleration represents A) B) C) D)

the time rate of change in the magnitude of the velocity. the time rate of change in the direction of the velocity. magnitude of the velocity. direction of the total acceleration.

2. A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing at a constant rate of 4 m/s2. What is the magnitude of its total acceleration at this instant? A) 3 m/s2 B) 4 m/s2 C) 5 m/s2 D) -5 m/s2 3. If a particle moving in a circular path of radius 5 m has a velocity function v = 4t2 m/s, what is the magnitude of its total acceleration at t = 1 s? A) 8 m/s B) 8.6 m/s C) 3.2 m/s

D) 11.2 m/s


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12-8. Curvilinear Motion: Cylindrical Components Application:

The boy slides down the slide at a constant speed of 2 m/s. Problem: How fast is his elevation from the ground changing (i.e., . what is z )? In general, the cylindrical coordinate system is used in cases where the particle moves along a 3-D curve.

Fig.12-13

Objective: To determine velocity and acceleration components using cylindrical coordinates.


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Chinese University of Hong Kong Special case:

Fig.12-14

9 A polar coordinate system is a 2-D representation of the cylindrical coordinate system. 9 When the particle moves in a plane (2-D), and the radial distance, r, is not constant, the polar coordinate system can be used to express the path of motion of the particle. If motion is restricted to the plane, polar coordinates r and θ are used Department of Mechanical and Automation Engineering

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Polar Coordinates: Position • Specify the location of P using both the radial coordinate r, which extends outward from the fixed origin O to the particle and a traverse coordinate θ, which is the counterclockwise angle between a fixed reference line and the r axis; • Angle usually measured in degrees or radians, where 1 rad = 180° • Positive directions of the r and θ coordinates are defined by the unit vectors ur and uθ • ur extends from P along increasing r, when θ is held fixed • uθ extends from P in the direction that occurs when r is held fixed and θ is increased • Note these directions are perpendicular to each other Department of Mechanical and Automation Engineering

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9 Position We can express the location of P in polar coordinates as r = r ur

(12-20)

Note that the radial direction, r, extends outward from the fixed origin, O, and the transverse coordinate, θ, is measured counter-clockwise (CCW) from the horizontal.


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9 Velocity

The instantaneous velocity is defined as:

v = r& = r&u r + ru& r Using the chain rule: dur/dt = (dur/dθ)(dθ/dt)

.

We can prove that dur/dθ = uθ so dur/dt = θ uθ

.

.

Therefore: v = r ur + rθ uθ Thus, the velocity vector has two components: r& , called the radial component, and rθ& , called the transverse component. The speed of the particle at any given instant is the sum of the squares of both components . .2 or (12-22) 2

v = (r) + ( rθ)

(12-21)

Fig.12-15(c)


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9 Acceleration

The instantaneous acceleration is defined as:

.

.

a = dv/dt = (d/dt)( rur + rθ uθ) After manipulation, the acceleration can be expressed as

. .. .. .. 2 a = (r – rθ )ur + (rθ + 2rθ)uθ (12-23) .. . 2 The term (r – rθ ) is the radial acceleration or ar. .. . . The term (rθ + 2rθ) is the transverse acceleration or aθ The magnitude of acceleration is

a=

(&r& − rθ ) + (rθ&& + 2r&θ&)2 2 2 &


(12-24) 27


Cylindrical Coordinates If the particle P moves along a space curve, its position can be written as

rP = rur + zuz Taking time derivatives and using the chain rule: Velocity:

v = r&u r + rθ& u θ + z&u z Fig.12-16

Acceleration:

a = (&r& − rθ& 2 )u r + (rθ&& + 2r&θ&)u θ + &z&u z


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Example 5 The rob OA is rotating in the horizontal plane such that θ = t3 rad. At the same time, the collar B is sliding outwards along OA so that r = (100t2)mm. If in both cases, t is in seconds, determine the velocity and acceleration of the collar when t = 1s. Solution:

Coordinate System. Since time-parametric equations of the particle is given, it is not necessary to relate r to θ. Velocity and Acceleration. r = 100t 2 r& = 200t &r& = 200

t =1s

t =1s t =1s

= 100mm θ = t 3

t =1s

= 200mm / s θ& = 3t 2 = 200mm / s 2 θ&& = 6t

= 1rad = 5.73° t =1s t =1s

= 2rad / s

Fig.12-17

= 6rad / s 2


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v = r&u r + rθ&u θ = {200u r + 300u θ } mm/s The magnitude of v is v = 200 2 + 300 2 = 361mm/s ⎛ 300 ⎞ o δ = tan −1 ⎜ ⎟ = 56.3 ⎝ 200 ⎠ a = (&r& − rθ& 2 )u r + (rθ&& + 2r&θ&)u θ = {−700u r + 1800u θ } mm/s 2

The magnitude of a is a = 700 2 + 1800 2 = 1930 mm/s 2 ⎛ 1800 ⎞ o φ = tan −1 ⎜ ⎟ = 68.7 ⎝ 700 ⎠


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.

Example 6 Given: r = 5 cos(2θ) m, θ = 3t2 (rad/s), θo = 0 Find: Velocity and acceleration at θ = 30°. Solution:

Apply chain rule to determine evaluate at θ = 30°.

θ =∫

t

t0 =0

r& and &r& and

t & θdt = ∫ 3t 2 dt = t 3 0

At θ = 30o , θ = π = t 3 Therefore: t=0.806 s. 6 θ& = 3t 2 = 3(0.806) 2 = 1.95 rad/s

θ&& = 6t = 6(0.806) = 4.836 rad/s 2

Fig.12-18

r = 5 cos(2θ ) = 5 cos(60 o ) = 2.5 m

r& = −10 sin( 2θ )θ& = −10 sin(60o )(1.95) = −16.88 m/s Department of Mechanical and Automation Engineering

.

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&r& = −20 cos(2θ )θ& 2 − 10 sin( 2θ )θ&& = −20 cos(60 o )(1.95) 2 − 10 sin(60 o )(4.836) = −80 m/s 2 Substitute in the equation for velocity

v = r&u r + rθ&u θ = −16.88u r + 2.5(1.95)u θ

v = (16.88) 2 + (4.87) 2 = 17.57 m/s Substitute in the equation for acceleration:

a = (&r& − rθ& 2 )u r + (rθ&& + 2r&θ&)u θ = [−80 − 2.5(1.95) 2 ]u r + [2.5(4.836) + 2(−16.88)(1.95)]u θ = −89.5u r − 53.7u θ m/s 2 a = (89.5) 2 + (53.7) 2 = 104.4 m/s 2 Department of Mechanical and Automation Engineering

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Exercise: 1. In a polar coordinate system, the velocity vector can be written as v = vr u r + vθ u θ = r&u r + rθ&u θ . The term θ& is called A) transverse velocity. C) angular velocity.

B) radial velocity. D) angular acceleration.

2. The speed of a particle in a cylindrical coordinate system is A)

r&

B) rθ&

C)

(rθ&) 2 + (r&) 2 D)

(rθ&) 2 + (r&) 2 + ( z& ) 2

3. If r& is zero for a particle, the particle is A) not moving. C) moving on a straight line.

B) moving in a circular path. D) moving with constant velocity.


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4. If a particle moves in a circular path with constant velocity, its radial acceleration is A) zero.

B)

&r&

.

C) − rθ& 2 . D) 2r&θ&

5. The radial component of velocity of a particle moving in a circular path is always A) zero. B) constant. C) greater than its transverse component. D) less than its transverse component. 6. The radial component of acceleration of a particle moving in a circular path is always A) negative. B) directed toward the center of the path. C) perpendicular to the transverse component of acceleration. D) All of the above. Department of Mechanical and Automation Engineering

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12-9. Absolute Dependent Motion Analysis of Two Particles Application:

Rope and pulley arrangements are often used to assist in lifting heavy objects. The total lifting force required from the truck depends on the acceleration of the cabinet. Problem: How can we determine the acceleration and velocity of the cabinet if the acceleration of the truck is known?

Fig.12-19

Objective: To relate the positions, velocities, and accelerations of particles undergoing dependent motion.


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Dependent Motion In many kinematics problems, the motion of one object will depend on the motion of another object. The blocks in this figure are connected by an inextensible cord wrapped around a pulley. If block A moves downward along the inclined plane, block B will move up the other incline. Fig.12-20

The motion of each block can be related mathematically by defining position coordinates, sA and sB. Each coordinate axis is defined from a fixed point or datum line, measured positive along each plane in the direction of motion of each block. Department of Mechanical and Automation Engineering

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9 In this example, position coordinates sA and sB can be defined from fixed datum lines extending from the center of the pulley along each incline to blocks A and B.

9 If the cord has a fixed length, the position coordinates sA and sB are related mathematically by the equation

sA + lCD + sB = lT where lT is the total cord length and lCD is the length of cord passing over arc CD on the pulley.


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9 The velocities of blocks A and B can be related by differentiating the position equation. Note that lCD and lT remain constant, so dlCD/dt = dlT/dt = 0

dsA/dt + dsB/dt = 0 =>

vB = -vA

9 The negative sign indicates that as A moves down the incline (positive sA direction), B moves up the incline (negative sB direction). 9 Accelerations can be found by differentiating the velocity expression. [Prove to yourself that aB = -aA] .


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Example

Position coordinates (sA and sB) are defined from fixed datum lines, measured along the direction of motion of each block. Note that sB is only defined to the center of the pulley above block B, since this block moves with the pulley. Also, h is a constant. The red colored segments of the cord remain constant in length during motion of the blocks. The position coordinates are related by the equation

2sB + h + sA = l where l is the total cord length minus the lengths of the red segments.

Fig.12-21


39


Since l and h remain constant during the motion, the velocities and accelerations can be related by two successive time derivatives:

2vB = -vA

and

2aB = -aA

When block B moves downward (+sB), block A moves to the left (-sA). Remember to be consistent with the sign convention!

This example can also be worked by defining the position coordinate for B (sB) from the bottom pulley instead of the top pulley. The position, velocity, and acceleration relations then become

2(h – sB) + h + sA = l and

2vB = vA

2aB = aA

[Prove to yourself that the results are the same, even if the sign conventions are different than the previous formulation.] Department of Mechanical and Automation Engineering

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Example 7 In the figure on the left, the cord at A is pulled down with a speed of 8 m/s. Find the speed of block B. Solution:

1) Define the position coordinates from a fixed datum line

DATUM

sA

sC

sB

Fig.12-22


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2). Define l1 as the length of the first cord, minus any segments of constant length. Define l2 in a similar manner for the second cord: Cord 1: 2sA + 2sC = l1 Cord 2: sB + (sB – sC) = l2 3) Eliminating sC between the two equations, we get 2sA + 4sB = l1 + 2l2 4) Relate velocities by differentiating this expression. Note that l1 and l2 are constant lengths. 2vA + 4vB = 0

=>

vB = - 0.5vA = - 0.5(8) = - 4 m/s

The velocity of block B is 4 m/s up (negative sB direction). Department of Mechanical and Automation Engineering

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Exercise: 1. Two blocks are interconnected by a cable. Which of the following is correct? A) vA= - vB C) (vy)A= - (vy)B

y

B) (vx)A= - (vx)B D) All of the above.

x

2. Determine the speed of block B when block A is moving down at 6 m/s while block C is moving down at 18 m/s . A) 24 m/s C) 12 m/s

B) 3 m/s D) 9 m/s

3. Determine the velocity vector of block A when block B is moving downward with a speed of 10 m/s. A) (8i + 6j) m/s C) (-8i - 6j) m/s

vA=6 m/s

vC=18 m/s

j

B) (4i + 3j) m/s D) (3i + 4j) m/s


vB=10 m/s

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12-10. Relative Motion Analysis of Two Particles Using Translating Axes Example:

When you try to hit a moving object, the position, velocity, and acceleration of the object must be known. Here, the boy on the ground is at d = 10ft when the girl in the window throws the ball to him. Problem: If the boy on the ground is running at a constant speed of 4 ft/s, how fast should the ball be thrown?

Fig.12-24

Objectives: a). To understand translating frames of reference. b). To use translating frames of reference to analyze Department relative of Mechanical and Automation Engineering motion. 44


9 Relative Position Consider particles A and B, which move along the arbitrary paths aa and bb, respectively, The absolute position of two particles A and B with respect to the fixed x, y, z reference frame are given by rA and rB. The position of B relative to A is represented by

rB/A = rB – rA

(12-25)

where rB/A is the relative-position vector which describes the relative position of “B with respect to A”

rB = (10i + 2j) m and then rB/A = (6i – 3j) m.

Example, if

Fig.12-25

rA = (4i + 5j) m,


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9 Relative Velocity To determine the relative velocity of B with respect to A, the time derivative of the relative position equation is taken. vB/A = vB – vA or vB = vA + vB/A (12-26) where vB and vA -- absolute velocities, since they are observed from the fixed frame. vB/A is the relative velocity of B w.r.t to A, because it is observed from the translating frame. Negative Note: vB/A = -vA/B

Fig.12-26

9 Relative Acceleration The time derivative of the relative velocity equation yields a similar vector relationship between the absolute and relative accelerations of particles A and B Fig.12-27 aB/A = aB – aA or aB = aA + aB/A (12-27) Department of Mechanical and Automation Engineering

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Example 8 At the instant, car A and B are traveling with the speed of 18 m/s and 12 m/s respectively. Also at this instant, A has a decrease in speed of 2 m/s2, and B has an increase in speed of 3 m/s2. Determine the velocity and acceleration of B with respect to A. Solution:

1). Velocity. The fixed x, y axes are established at a point on the ground and the translating x’, y’ axes are attached to car A. Using Cartesian vector analysis,

vB = v A + vB/ A

(

)

− 12 j = − 18 cos 60o i − 18 sin 60o j + v B / A v B / A = {9i + 3.588 j} m / s

Fig.12-28 Department of Mechanical and Automation Engineering

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Thus,

vB / A = 9 2 + 3.5882 = 9.69 m / s

Its direction is

(vB / A )y tan θ = (vB / A )x

=

3.588 , ⇒ θ = 21.7 o 9

2). Acceleration. The magnitude of the normal component is

(aB )n =

Applying the equation for relative acceleration yields

vB2

ρ

= 1.440 m / s 2

aB = a A + aB/ A

(− 1.440i − 3j) = (2 cos 60o i + 2 sin 60o j) + a B / A a B / A = {− 2.440i − 4.732 j} m / s 2 Magnitude and direction are

aB / A = 5.32 m / s 2 , ⇒ φ = 62.7 o Department of Mechanical and Automation Engineering

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Exercise: 1. The velocity of B relative to A is defined as A) vB – vA . B) vA – vB . C) vB + vA . D) vA + vB . 2. Since vector addition forms a triangle, there can be at most _________ unknowns (either magnitudes and/or directions of the vectors) to solve the problem. m A) one B) two C) three D) four vB = 4 s 3. Two particles, A and B, are moving in the directions shown. What should be the angle θ so that vB/A is minimum? A) 0° B) 180°

C) 90°

D) 270°

B

θ

A

vA = 3 ms

4. Determine the velocity of plane A with respect to plane B. A) (400i+520j) km/hr B) (1220i-300j) km/hr 30 C) (-181i-300j) km/hr ° D) (-1220i+300j) km/hr


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2. Curvilinear Motion.pdf

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